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Harman [31]
3 years ago
13

Of the 800 seventh grade students and 640 eighth grade students at ​Knox Middle School, 90 students will be chosen at random to

attend an ​assembly.
​
​Which of the following would be a representative sample of ​the number of seventh and eighth grade students at the school?
Mathematics
1 answer:
Irina18 [472]3 years ago
3 0

Answer:

50 7th grade and 40 8th grade

Step-by-step explanation:

1. add to get total number of students, 1440

2. find percentage of total that are 7th grade students and 8th grade student:

640/1440 = ~45% (are 7th grade)

800/1440 = ~55% (are 8th grade)

3. 55% of 90 is 50, and 45% of 90 is 40

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Simplify: <br>{(12)^1 + (13)^-1}/[(1/5)^-2 × {(1/5)^-1 + (1/8)^-1}^-1]​
CaHeK987 [17]

Step-by-step explanation:

\underline{\underline{\sf{➤\:\:Solution}}}

\sf \dashrightarrow \:  \dfrac{ \left(\left(12 \right)^{ - 1}  + \left(13 \right)^{ - 1}  \right) }{\left( \dfrac{1}{5}\right) ^{ - 2}  \times\left( \left( \dfrac{1}{5}  \right) ^{ - 1}  +\left( \dfrac{1}{8}  \right) ^{ - 1}  \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1}{12}  + \dfrac{1}{13} \right) }{\left( \dfrac{5}{1}\right) ^{ 2}  \times\left( \dfrac{5}{1}  + \dfrac{8}{1}   \right) ^{ - 1}}

  • LCM of 12 and 13 is 156

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{1 \times 13 = 13}{12 \times 13 = 156}  + \dfrac{1 \times 12 = 12}{13 \times 12 = 156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{5 + 8}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13}{156}  + \dfrac{12}{156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{13}{1}    \right) ^{ - 1}}

\sf \dashrightarrow \:  \dfrac{ \left(\dfrac{13 + 12}{156}  \right) }{\ \dfrac{25}{1} \times\dfrac{1}{13} }

\sf \dashrightarrow \:    \dfrac{25}{156} \div    \dfrac{25}{13}

\sf \dashrightarrow \:    \dfrac{ \cancel{25}}{156}  \times    \dfrac{13}{ \cancel{25} }

\sf \dashrightarrow \:     \dfrac{13}{156}

\sf \dashrightarrow \:     \dfrac{1}{12}

\sf \dashrightarrow \:    Answer =   \underline{\boxed{ \sf{ \dfrac{1}{12} }}}

━━━━━━━━━━━━━━━━━━━━━━━━

\underline{\underline{\sf{★\:\:Laws\:of\: Exponents :}}}

\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}

\sf 2^{nd} \: Law =

\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}

\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}

\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}

\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}

\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1

3 0
2 years ago
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astraxan [27]
I think it is the fourth one
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On 1, I need help, and for 2, is the answer right?
kramer

Answer:

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Step-by-step explanation:

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5 0
2 years ago
8(s+3)=72 i need help it is due tomorrow
Montano1993 [528]

Answer:

s = 6

Step-by-step explanation:

8(s+3) = 72

Expand the brackets out:

(8 x s) + (8 x 3) = 72

8s + 24 = 72

Now you get rid of 24 on the left, by taking away 24. But you also have to do the same on the right side so that both sides are equal:

8s + 24 = 72

left(-24) = right (-24)

So that leaves you with:

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Now divide both sides by 8 to make them equal and to get s on its own:

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s = 6


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Step-by-step explanation:

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